Question

Let T1 and T2 be two distinct common tangents to the ellipse $E: \frac{x^2}{6} + \frac{y^2}{3} = 1$ and the parabola $P: y^2 = 12x$. Suppose that the tangent T1 touches P and E at the points A1 and A2, respectively and the tangent T2 touches P and E at the points A4 and A3, respectively. Then which of the following statements is(are) true?

• A. The area of the quadrilateral A1A2A3A4 is 35 square units
• B. The area of the quadrilateral A1A2A3A4 is 36 square units
• C. The tangents T1 and T2 meet the x -axis at the point (–3,0)
• D. The tangents T1 and T2 meet the x -axis at the point (–6,0)

Answer 1

Answer: (A), (C)

The area of the quadrilateral A1A2A3A4 is 35 square units

Answer 2

Given :
$E:\frac{x^2}{6}+\frac{y^2}{3}=1$, Tangent : $y=m_1x±\sqrt{6m_1^2+3}$
P : y2 = 12x, Tangent : $y=m_2x+\frac{3}{m_2}$
Now, for common tangent :
$m=m_1+m_2,±\sqrt{6m_1^2+3}=\frac{3}{m_2}$
⇒ m = ±1
The equations of the common tangents are y = x + 3 and y = -x - 3. The points of contact for the parabola are:
$\left(\frac{a}{m^2},\frac{2a}{3}\right)$
A1 ≡ (3, 6), A4(3 - 6)
Now, Let suppose A2(x1, y1)
⇒ tangent to E : $\frac{xx_1}{6}+\frac{yy_1}{3}=1$
A3 is mirror image of A2 in x-axis ⇒ A3(–2, –1)

The intersection point of T1 = 0 and T2 = 0 is at (-3, 0).
Area of quadrilateral A1A2A3A4 :
$\frac{1}{2}(12+2)\times5=35$ square units.
So, the correct options are (A) and (C).